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Evaluate the limit for the following if exists.$\;\lim\limits_{x\to 2} \large\frac{\sin \pi x}{2 - x}$

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  • L'Hopital's rule: Let $f$ and $g$ be continous real valued functions defined on the closed interval $[a,b], f,g$ be differentiable on $(a,b)$ and $g'(c) \neq 0$
  • Then if $ \lim\limits_{x \to c}\; f(x)=0, \lim \limits_{x \to c}\; g(x)=0$ and
  • $ \lim\limits_{x \to c} \large\frac{f'(x)}{g'(x)}$$=L$ it follows that
  • $ \lim \limits_{x \to c} \large\frac{f(x)}{g(x)}$$=L$
Step 1:
.$\;\lim\limits_{x\to 2} \large\frac{\sin \pi x}{2 - x}$ is of the form $\large\frac{0}{0}$
Step 2:
Applying L'Hopital's rule,
$\;\lim\limits_{x\to 2} \large\frac{\sin \pi x}{2 - x}$$=\lim\limits_{x\to 2} \large\frac{\pi \cos \pi x}{-1}$$=-\pi$
answered Jul 29, 2013 by meena.p

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